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Math-6139-A1 · revision r1 · on Math-6139

Reused writeup on Math-6139 (gpt pro 5.2)

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem 1085 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. Erdős Problem 1085 (unit distances without separation)
  5. FORMAL RESTATEMENT
  6. QUICK LITERATURE/CONTEXT CHECK
  7. ATTACK PLAN
  8. WORK
  9. Fast reality check.
  10. VERIFICATION
  11. FINAL
  12. COMPLETION ESTIMATE

Erdős Problem 1085 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1085

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1085.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1085.tex.

Erdős Problem 1085 (unit distances without separation)

FORMAL RESTATEMENT

For integers $d\ge 1$ and $n\in\mathbb{N}$, let $f_d(n)$ be the minimum integer such that every set $P\subset\mathbb{R}^d$ with $|P|=n$ determines at most $f_d(n)$ unordered pairs $\{x,y\}\subset P$ with $\|x-y\|=1$. Equivalently, $f_d(n)=\max\{\#\{\{x,y\}\subset P:\|x-y\|=1\}: P\subset\mathbb{R}^d,\ |P|=n\}$. The problem asks for estimates of $f_d(n)$, especially for $d=2,3$.

QUICK LITERATURE/CONTEXT CHECK

The problem statement records the best-known bounds in $d=2$ and $d=3$ and the asymptotically tight quadratic behaviour for $d\ge 4$ (including the Lenz construction and an Erdős-Stone based upper bound for $d\ge 4$). I do not add other literature.

ATTACK PLAN

  • Proof track (partial). Give a completely self-contained quadratic lower bound construction in $d\ge 4$ (Lenz-type: points on mutually orthogonal circles), and a general upper bound from forbidding large equidistant cliques plus Turán.
  • Disproof track. Not applicable (estimation problem).

WORK

Fast reality check.
For $d=4$, taking two circles in orthogonal planes of radius $1/\sqrt2$ and placing $m$ points on each gives $n=2m$ points with exactly $m^2$ cross pairs at distance $1$. Choosing $m=3$ equally spaced points on each circle, a brute-force check gives $9$ cross pairs and the set of cross squared distances is exactly $\{1\}$ (within-circle squared distances are $\{3/2\}$).
Lemma (Maximum size of an equidistant set): In $\mathbb{R}^d$, there do not exist $d+2$ points with all pairwise distances equal to $1$. Equivalently, any clique in a unit-distance graph in $\mathbb{R}^d$ has size at most $d+1$.
Proof: Suppose for contradiction that $p_1,\dots,p_{d+2}\in\mathbb{R}^d$ satisfy $\|p_i-p_j\|=1$ for all $i\ne j$. Translate so that $p_{d+2}=0$. Then $\|p_i\|=1$ for $i\le d+1$, and for $1\le i<j\le d+1$, \[ 1=\|p_i-p_j\|^2=\|p_i\|^2+\|p_j\|^2-2p_i\cdot p_j=1+1-2p_i\cdot p_j, \] so $p_i\cdot p_j=1/2$. Let $G$ be the $(d+1)\times(d+1)$ Gram matrix $G=(p_i\cdot p_j)_{1\le i,j\le d+1}$. Then $G$ has diagonal entries $1$ and off-diagonal entries $1/2$, so \[ G=\frac12 I_{d+1}+\frac12 J_{d+1}, \] where $J_{d+1}$ is the all-ones matrix. The eigenvalues of $J_{d+1}$ are $d+1$ (once) and $0$ (multiplicity $d$), hence the eigenvalues of $G$ are $\tfrac12+\tfrac12(d+1)=\tfrac{d+2}{2}$ (once) and $\tfrac12$ (multiplicity $d$). Thus $\mathrm{rank}(G)=d+1$. But $\mathrm{rank}(G)\le d$ because $G$ is the Gram matrix of $d+1$ vectors in $\mathbb{R}^d$. Contradiction. Therefore no such $d+2$ points exist. ∎
Lemma (A general Turán-type upper bound): For any $d\ge 1$ and $n$, \[ f_d(n)\le \left(1-\frac{1}{d+1}\right)\frac{n^2}{2}. \]
Proof: Given $P\subset\mathbb{R}^d$ of size $n$, let $G$ be its unit-distance graph. By the previous lemma, $G$ is $K_{d+2}$-free. Turán's theorem for $K_{r}$-free graphs states that a $K_r$-free graph on $n$ vertices has at most $\left(1-\frac{1}{r-1}\right)\frac{n^2}{2}$ edges. Taking $r=d+2$ yields $|E(G)|\le \left(1-\frac{1}{d+1}\right)\frac{n^2}{2}$. Since $f_d(n)$ is the maximum possible $|E(G)|$, the same bound holds for $f_d(n)$. ∎
Lemma (Lenz-type construction for $d\ge 4$): Let $d\ge 4$, and set $p:=\lfloor d/2\rfloor\ge 2$. Then for all $n$ there exists a set $P\subset\mathbb{R}^d$ of $|P|=n$ points with \[ \#\{\{x,y\}\subset P: \|x-y\|=1\}\ \ge\ \frac{p-1}{2p}n^2 - O(p). \] In particular, $f_d(n)\ge \frac{p-1}{2p}n^2-O(1)$.
Proof: Embed $p$ pairwise orthogonal 2-dimensional linear subspaces $V_1,\dots,V_p\subset\mathbb{R}^d$ (possible since $2p\le d$). In each $V_i$, consider the circle $C_i:=\{x\in V_i: \|x\|=1/\sqrt2\}$, centered at the origin. For any $x\in C_i$ and $y\in C_j$ with $i\ne j$, orthogonality gives $x\cdot y=0$, hence \[ \|x-y\|^2=\|x\|^2+\|y\|^2-2x\cdot y = \frac12+\frac12-0=1. \] Thus every pair of points chosen from two different circles is at unit distance.

Now distribute $n$ points among the circles: choose integers $n_1,\dots,n_p\ge 0$ with $\sum_i n_i=n$, and pick any $n_i$ distinct points on each $C_i$. The number of unit-distance pairs between different circles equals $\sum_{1\le i<j\le p} n_i n_j = \frac12\left(n^2-\sum_{i=1}^p n_i^2\right)$. This is maximized when the $n_i$ are as equal as possible, giving $\sum_i n_i^2\le \lceil n/p\rceil^2+(p-1)\lfloor n/p\rfloor^2\le n^2/p + O(p)$. Therefore \[ \sum_{i<j}n_i n_j \ge \frac12\left(n^2-\frac{n^2}{p}\right)-O(p)=\frac{p-1}{2p}n^2-O(p). \] Any additional unit pairs within a single circle can only increase the count, so this lower bound holds for the full unit-distance count. ∎

VERIFICATION

  • The equidistant-set lemma is a standard rank/Gram-matrix argument; all eigenvalues computed explicitly.
  • The Lenz construction uses only orthogonality of subspaces and radii $1/\sqrt2$; the distance computation is exact.
  • The Turán bound is not sharp for $d\ge 4$ compared to the conjectured/recorded asymptotic $(p-1)/(2p)$, but it is rigorous and self-contained.

FINAL

UNRESOLVED.
  1. (i) Strongest proved partial result here: for $d\ge 4$ with $p=\lfloor d/2\rfloor$, $\frac{p-1}{2p}n^2-O(1)\le f_d(n)\le \left(1-\frac{1}{d+1}\right)\frac{n^2}{2}$.
  2. (ii) First gap: for $d=2,3$, close the gap between known lower and upper bounds (as recorded in the problem statement). For $d\ge 4$, prove the sharp asymptotic upper constant $(p-1)/(2p)$ without appealing to results not already in the file.
  3. (iii) Top 3 next moves: $1$ understand which complete multipartite graphs are forbidden in unit-distance graphs in $\mathbb{R}^d$; $2$ attempt to show near-extremal graphs must be approximately $p$-partite, matching the Lenz construction; $3$ computational experiments in $d=4$ to see extremal configurations for moderate $n$ beyond the circle construction.
  4. (iv) Minimal counterexample structure (to the sharp quadratic constant in $d\ge 4$): a unit-distance graph on $n$ vertices with edge density exceeding $(p-1)/(2p)+\varepsilon$, which would necessarily force a rich family of geometric constraints (many large cliques/near-cliques) contradicting Euclidean realizability.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.